Question: The equation of a circle $C$ is $x^2+y^2-16x-10y+40 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Solution: To find the equation in standard form, complete the square. $(x^2-16x) + (y^2-10y) = -40$ $(x^2-16x+64) + (y^2-10y+25) = -40 + 64 + 25$ $(x-8)^{2} + (y-5)^{2} = 49 = 7^2$ Thus, $(h, k) = (8, 5)$ and $r = 7$.